JAMB - Physics (1997 - No. 10)
An electric water pump rate 1.5 kW, lifts 200kg of water through a vertical height of 6 meters in 10 seconds. What is the efficiency of the pump? [g = 10ms-2, neglecting air resistance]
90.0%
85.0%
80.0%
65.0%
Explanation
Efficiency = output x 100
= \(\frac{\frac{200 {\times} 10 {\times} {6}}{10}}{1500}\) x \(\frac{100}{1}\)
= 80%
= \(\frac{\frac{200 {\times} 10 {\times} {6}}{10}}{1500}\) x \(\frac{100}{1}\)
= 80%
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